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You have 6 sticks of lengths 10, 20, 30, 40, 50, and 60 [#permalink]
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Updated on: 02 Jun 2012, 12:45
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43% (01:39) correct 57% (02:02) wrong based on 198 sessions
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You have 6 sticks of lengths 10, 20, 30, 40, 50, and 60 centimeters. The number of noncongruent triangles that can be formed by choosing three of the sticks to make the sides is A. 3 B. 6 C. 7 D. 10 E. 12 OA will be posted after some time. Please inclease my Kudos if you like the problem...
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Originally posted by sandal85 on 02 Jun 2012, 09:22.
Last edited by Bunuel on 02 Jun 2012, 12:45, edited 1 time in total.
Edited the question and added the OA



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Re: You have 6 sticks of lengths 10, 20, 30, 40, 50, and 60 [#permalink]
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02 Jun 2012, 12:49
sandal85 wrote: You have 6 sticks of lengths 10, 20, 30, 40, 50, and 60 centimeters. The number of noncongruent triangles that can be formed by choosing three of the sticks to make the sides is
A. 3 B. 6 C. 7 D. 10 E. 12
OA will be posted after some time.
Please inclease my Kudos if you like the problem... The length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides.Based on this there can be only 7 triangles formed: (20, 30, 40), (20, 40, 50), (20, 50, 60), (30, 40, 50), (30, 40, 60), (30, 50, 60), (40, 50, 60). Answer; C.
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Re: You have 6 sticks of lengths 10, 20, 30, 40, 50, and 60 [#permalink]
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07 Sep 2012, 12:12
Bunuel wrote: sandal85 wrote: You have 6 sticks of lengths 10, 20, 30, 40, 50, and 60 centimeters. The number of noncongruent triangles that can be formed by choosing three of the sticks to make the sides is
A. 3 B. 6 C. 7 D. 10 E. 12
OA will be posted after some time.
Please inclease my Kudos if you like the problem... The length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides.Based on this there can be only 7 triangles formed: (20, 30, 40), (20, 40, 50), (20, 50, 60), (30, 40, 50), (30, 40, 60), (30, 50, 60), (40, 50, 60). Answer; C. Hi Bunuel, Is there any other method (combinatorics) to solve this question ?
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Re: You have 6 sticks of lengths 10, 20, 30, 40, 50, and 60 [#permalink]
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09 Sep 2012, 04:32
fameatop wrote: Hi Bunuel,
Is there any other method (combinatorics) to solve this question ?
Following method may be applied As Bunuel has mentioned "The length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides."This a very important characteristic of triangle. [Thanks Bunuel, you are genius !!! ] Now coming to this particular problem, we have sides mentioned as 10, 20, 30, 40, 50, 60. Difference between any tow sides is 10, therefore 10 cannot be any side of a triangle. So, we are left with 20, 30, 40, 50 and 60. >> 5 sides. Now, Number of triangles can be formed with 5 sides is \(^^5C {^}3\) \(= 10\) However, following triangles are not possible \([ 20, 30, 50 ]  as 20 + 30 (NOT) > 50\) Similarly \([20, 30, 60 ]\) is not possible Similarly \([20, 40, 60]\) is not possible [Note: Easiest way to eliminate triangles is to start with least two sides and sum them up. All the sides having equal or higher that value are eliminated. In our case we started with 20 and 30. Hence, 50, and 60 are eliminated. Repeat the same process with the next number along with the least one. In our case 20 and 40. So, 40 is eliminated. ]So, finally the number of triangles is \(103 =7\) Hope, this is more logical and quicker method of solving such problem.
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Re: You have 6 sticks of lengths 10, 20, 30, 40, 50, and 60 [#permalink]
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07 Dec 2015, 01:53
Got this correct but spent 3.48 Used trial and error, keeping the basic properties of triangle in mind. Anybody has a better / faster method?? Thanks.



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Re: You have 6 sticks of lengths 10, 20, 30, 40, 50, and 60 [#permalink]
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10 Jun 2018, 06:55
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Re: You have 6 sticks of lengths 10, 20, 30, 40, 50, and 60
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10 Jun 2018, 06:55






